The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.4$ years; the standard deviation is $1.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $16.6$ years.
$13.4$ $11.8$ $15$ $10.2$ $16.6$ $8.6$ $18.2$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $13.4$ years. We know the standard deviation is $1.6$ years, so one standard deviation below the mean is $11.8$ years and one standard deviation above the mean is $15$ years. Two standard deviations below the mean is $10.2$ years and two standard deviations above the mean is $16.6$ years. Three standard deviations below the mean is $8.6$ years and three standard deviations above the mean is $18.2$ years. We are interested in the probability of a lion living less than $16.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $10.2$ years and the other half $({2.5\%})$ will live longer than $16.6$ years. The probability of a particular lion living less than $16.6$ years is ${95\%} + {2.5\%}$, or $97.5\%$.